NCERT Solutions for Exercise 6.2 Class 10 Maths Chapter 6 - Triangles

# NCERT Solutions for Exercise 6.2 Class 10 Maths Chapter 6 - Triangles

Edited By Ramraj Saini | Updated on Nov 17, 2023 05:21 PM IST | #CBSE Class 10th

## NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2

NCERT Solutions for Exercise 6.2 Class 10 Maths Chapter 6 Triangles are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. Class 10 maths ex 6.2 contains two theorems: (Theorem 1) When a parallel line is drawn along one of the triangle's sides and intersects the other two sides at certain points, the two sides are divided in the same proportion. (Theorem 2) Any line that divides the two sides of a triangle in the same ratio is parallel to the third side.

NCERT solutions for exercise 6.2 Class 10 Maths chapter 6 Triangles discusses about the concept triangle as a polygon, interior angles of a triangle, exterior angles of triangle, Pythagoras' theorem, and conditions of similarity in triangles. 10th class Maths exercise 6.2 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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## Assess NCERT Solutions for Class 10 Maths chapter 6 exercise 6.2

Triangles Class 10 Chapter 6 Exercise: 6.2

(i)

Let EC be x

Given: DE || BC

By using the proportionality theorem, we get

$\frac{AD}{DB}=\frac{AE}{EC}$

$\Rightarrow \frac{1.5}{3}=\frac{1}{x}$

$\Rightarrow x=\frac{3}{1.5}=2\, cm$

$\therefore EC=2\, cm$

(ii)

Given: DE || BC

By using the proportionality theorem, we get

$\frac{AD}{DB}=\frac{AE}{EC}$

$\Rightarrow \frac{x}{7.2}=\frac{1.8}{5.4}$

$\Rightarrow x=\frac{7.2}{3}=2.4\, cm$

$\therefore AD=2.4\, cm$

(i)

Given :

PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

$\frac{PE}{EQ}=\frac{3.9}{3}=1.3\, cm$ and $\frac{PF}{FR}=\frac{3.6}{2.4}=1.5\, cm$

We have

$\frac{PE}{EQ} \neq \frac{PF}{FR}$

Hence, EF is not parallel to QR.

(ii)

Given :

PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm

$\frac{PE}{EQ}=\frac{4}{4.5}=\frac{8}{9}\, cm$ and $\frac{PF}{FR}=\frac{8}{9}\, cm$

We have

$\frac{PE}{EQ} = \frac{PF}{FR}$

Hence, EF is parallel to QR.

(iii)

Given :

PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

$\frac{PE}{PQ}=\frac{0.18}{1.28}=\frac{9}{64}\, cm$ and $\frac{PF}{PR}=\frac{0.36}{2.56}=\frac{9}{64}\, cm$

We have

$\frac{PE}{EQ} = \frac{PF}{FR}$

Hence, EF is parallel to QR.

Given : LM || CB and LN || CD

To prove :

$\frac{AM}{AB} = \frac{AN}{AD }$

Since , LM || CB so we have

$\frac{AM}{AB}=\frac{AL}{AC}.............................................1$

Also, LN || CD

$\frac{AL}{AC}=\frac{AN}{AD}.............................................2$

From equation 1 and 2, we have

$\frac{AM}{AB} = \frac{AN}{AD }$

Hence proved.

Given : DE || AC and DF || AE.

To prove :

$\frac{BF}{FE} = \frac{BE}{EC }$

Since , DE || AC so we have

$\frac{BD}{DA}=\frac{BE}{EC}.............................................1$

Also,DF || AE

$\frac{BD}{DA}=\frac{BF}{FE}.............................................2$

From equation 1 and 2, we have

$\frac{BF}{FE} = \frac{BE}{EC }$

Hence proved.

Given : DE || OQ and DF || OR.

To prove EF || QR.

Since DE || OQ so we have

$\frac{PE}{EQ}=\frac{PD}{DO}.............................................1$

Also, DF || OR

$\frac{PF}{FR}=\frac{PD}{DO}.............................................2$

From equation 1 and 2, we have

$\frac{PE}{EQ} = \frac{PF}{FR }$

Thus, EF || QR. (converse of basic proportionality theorem)

Hence proved.

Given : AB || PQ and AC || PR

To prove: BC || QR

Since, AB || PQ so we have

$\frac{OA}{AP}=\frac{OB}{BQ}.............................................1$

Also, AC || PR

$\frac{OA}{AP}=\frac{OC}{CR}.............................................2$

From equation 1 and 2, we have

$\frac{OB}{BQ} = \frac{OC}{CR }$

Therefore, BC || QR. (converse basic proportionality theorem)

Hence proved.

Let PQ is a line passing through the midpoint of line AB and parallel to line BC intersecting line AC at point Q.

i.e. $PQ||BC$ and $AP=PB$ .

Using basic proportionality theorem, we have

$\frac{AP}{PB}=\frac{AQ}{QC}..........................1$

Since $AP=PB$

$\frac{AQ}{QC}=\frac{1}{1}$

$\Rightarrow AQ=QC$

$\therefore$ Q is the midpoint of AC.

Let P is the midpoint of line AB and Q is the midpoint of line AC.

PQ is the line joining midpoints P and Q of line AB and AC, respectively.

i.e. $AQ=QC$ and $AP=PB$ .

we have,

$\frac{AP}{PB}=\frac{1}{1}..........................1$

$\frac{AQ}{QC}=\frac{1}{1}...................................2$

From equation 1 and 2, we get

$\frac{AQ}{QC}=\frac{AP}{PB}$

$\therefore$ By basic proportionality theorem, we have $PQ||BC$

Draw a line EF passing through point O such that $EO||CD\, \, and\, \, FO||CD$

To prove :

$\frac{AO}{BO} = \frac{CO}{DO}$

In $\triangle ADC$ , we have $CD||EO$

So, by using basic proportionality theorem,

$\frac{AE}{ED}=\frac{AO}{OC}........................................1$

In $\triangle ABD$ , we have $AB||EO$

So, by using basic proportionality theorem,

$\frac{DE}{EA}=\frac{OD}{BO}........................................2$

Using equation 1 and 2, we get

$\frac{AO}{OC}=\frac{BO}{OD}$

$\Rightarrow \frac{AO}{BO} = \frac{CO}{DO}$

Hence proved.

Draw a line EF passing through point O such that $EO||AB$

Given :

$\frac{AO}{BO} = \frac{CO}{DO}$

In $\triangle ABD$ , we have $AB||EO$

So, by using basic proportionality theorem,

$\frac{AE}{ED}=\frac{BO}{DO}........................................1$

However, its is given that

$\frac{AO}{CO} = \frac{BO}{DO}..............................2$

Using equation 1 and 2 , we get

$\frac{AE}{ED}=\frac{AO}{CO}$

$\Rightarrow EO||CD$ (By basic proportionality theorem)

$\Rightarrow AB||EO||CD$

$\Rightarrow AB||CD$

Therefore, ABCD is a trapezium.

## More About NCERT Solutions for Class 10 Maths Exercise 6.2

NCERT solutions Class 10 Maths chapter 6 exercise 6.2 triangles necessitates a thorough understanding of the underlying theorems. As a result, students are advised to read through the logical explanations of the theorems' proofs, as well as practise them using well-explained diagrams.

NCERT solutions Class 10 Maths chapter 6 exercise 6.2 may be mastered if students remember and comprehend how the theorems are applied. Children should consider creating a theory chart that they can refer to from time to time. Another creative and simple way to comprehend theorems is to use everyday objects. Students might ask their teachers and parents for assistance in doing the same. Students can also access Triangles Class 10 Notes here and use them for quickly revision of the concepts related to Triangles.

## Benefits of NCERT Solutions for Class 10 Maths Exercise 6.2

• The NCERT solutions for Class 10 Maths exercise 6.2 guide students through the process of resolving and revising all of the exercise's questions.
• You will be able to achieve better scores if you look over the NCERT solution for class 10 Mathematics chapter 6 exercise 6.2 and practise it thoroughly.
• Congruent and related triangles and theorems are used in Class 10 Maths exercise 6.2.

Also see-

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## Subject Wise NCERT Exemplar Solutions

1. What are the figures that are similar?

The term "similar" refers to two figures that have the same shape but differ in size.

2. Difference between congruent and similar figures?

The figures that are congruent are all similar, but the opposite is not true.

3. What is the use of side ratios for similar triangle?

If we know the side ratio and the length of the any side, we can find the length of the corresponding side of the other triangle by using the ratio.

4. What is the ratio of the area of triangles if the ratio of the sides is given?

The ration of area of the triangles is equal to the square of the ratio of the sides.

5. If the ratio of the sides of two similar triangle is 1:2 then the ratio of their areas is?

The ration of area of the triangles is equal to the square of the ratio of the sides.

Ratio of area = 1:4

6. If the ratio of the sides is given as 1:3 and the shorter side has length of 4cm then the length of the longer side is?

4/x=1/3

x=12cm

The length of the longer side is 12cm.

7. Mention one of the properties of the line that passes through the mid-point of any two sides of the triangle?

It will always be parallel to the third side of the triangle.

8. Is similarity a sign of agreement between the figures?

No, resemblance only relates to the shape of the figures and does not imply equality.

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